Answer these questions ### Statistical Hypothesis Testing: Mean Tuna ConsumptionA nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 60 people shows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation is 1.24 pounds. At \( \alpha = 0.06 \), can you reject the claim?#### Step-by-Step Solution1. **Calculate the Test Statistic:** \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \] Given: - \( \bar{x} = 3.1 \) - \( \mu = 3.3 \) - \( \sigma = 1.24 \) - \( n = 60 \) \[ z = \frac{3.1 - 3.3}{1.24 / \sqrt{60}} \approx -1.25 \quad \text{(Rounded to two decimal places as needed.)} \]2. **Find the P-value:** Using the standard normal distribution table (Z-table), we find the area to the left of \( z = -1.25 \). \[ \text{P-value} = 0.211 \quad \text{(Rounded to three decimal places as needed.)} \]3. **Decision Rule:** Compare the P-value with the significance level \( \alpha = 0.06 \). - If \( \text{P-value} \leq \alpha \), reject the null hypothesis \( H_0 \). - If \( \text{P-value} > \alpha \), fail to reject the null hypothesis \( H_0 \). Here, \( \text{P-value} = 0.211 \) which is greater than \( \alpha = 0.06 \).4. **Conclusion:** - Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds.### Multiple Choice Decision#### Decide whether to reject or fail to reject the null hypothesis:- **A. Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds

It is given that Population mean = 3.3Sample mean = 3.1Population standard deviation = 1.24Sample…

hows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation 1.24 pounds. At a=0.06, can you reject the claim? =-1.25 (Round to two decimal places as needed.) c) Find the P-value. www 0.211 (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. OA. Fail to reject Ho. There is not sufficient evidence B. to reject the claim that mean tuna consumption is equal to 3.3 pounds. OC. Fail to reject Ho. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds. OD. 19 Reject Ho. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds. Reject Ho. There sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds.

hows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation 1.24 pounds. At a=0.06, can you reject the claim? =-1.25 (Round to two decimal places as needed.) c) Find the P-value. www 0.211 (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. OA. Fail to reject Ho. There is not sufficient evidence B. to reject the claim that mean tuna consumption is equal to 3.3 pounds. OC. Fail to reject Ho. There is sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds. OD. 19 Reject Ho. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds. Reject Ho. There sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Answer these questions

### Statistical Hypothesis Testing: Mean Tuna Consumption

A nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 60 people shows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation is 1.24 pounds. At \( \alpha = 0.06 \), can you reject the claim?

#### Step-by-Step Solution

1. **Calculate the Test Statistic:**
\[
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
\]
Given:
- \( \bar{x} = 3.1 \)
- \( \mu = 3.3 \)
- \( \sigma = 1.24 \)
- \( n = 60 \)

\[
z = \frac{3.1 - 3.3}{1.24 / \sqrt{60}} \approx -1.25 \quad \text{(Rounded to two decimal places as needed.)}
\]

2. **Find the P-value:**
Using the standard normal distribution table (Z-table), we find the area to the left of \( z = -1.25 \).

\[
\text{P-value} = 0.211 \quad \text{(Rounded to three decimal places as needed.)}
\]

3. **Decision Rule:**
Compare the P-value with the significance level \( \alpha = 0.06 \).

- If \( \text{P-value} \leq \alpha \), reject the null hypothesis \( H_0 \).
- If \( \text{P-value} > \alpha \), fail to reject the null hypothesis \( H_0 \).

Here, \( \text{P-value} = 0.211 \) which is greater than \( \alpha = 0.06 \).

4. **Conclusion:**
- Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds.

### Multiple Choice Decision

#### Decide whether to reject or fail to reject the null hypothesis:
- **A. Fail to reject \( H_0 \). There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds

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